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Proof step 24 from proof weakmon-lem (previous step)

∀ n. ii(n) ≤ ii(n + 1), ∀ n0, ii. n0 < m0 + 1 → (∀ n. ii(n) ≤ ii(n + 1)) → (∃ m. n0 ≤ ii(m)) ∨ (∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)) ⊦ ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n), ∃ m1. m0 < ii(m1), ¬ m0 < ii(m) ∧ (∃ m. ii(m) = m0)

User applied rule exists right

And obtained the following new goals

Proof step 25:
∀ n. ii(n) ≤ ii(n + 1), ∀ n0, ii. n0 < m0 + 1 → (∀ n. ii(n) ≤ ii(n + 1)) → (∃ m. n0 ≤ ii(m)) ∨ (∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n)) ⊦ ∃ n. (∀ m. ii(m) ≤ n) ∧ (∃ m. ii(m) = n), m0 < ii(m), ∃ m1. m0 < ii(m1), ¬ m0 < ii(m) ∧ (∃ m. ii(m) = m0)

Rule argument

pos: right 2 terms: m abort: F discard: F computed: F

Additional information

Used terms: m